Question

Let X be a normal random variable with X ? N (0, 1), what is the distribution function and density function of Y = X2

Answer 1

With
`Y=X^2`, note that the support of
`Y` will be all non-negative real numbers.


`F_Y(y)=\mathbb P(Y\le y)`

`F_Y(y)=\mathbb P(X^2\le y)`

`F_Y(y)=\mathbb P(|X|\le\sqrt y)`

`F_Y(y)=\mathbb P(-\sqrt y\le X\le\sqrt y)`

`F_Y(t)=\mathbb P(X\le\sqrt y)-\mathbb P(X\le-\sqrt y)`

`F_Y(y)=F_X(\sqrt y)-F_X(-\sqrt y)`

Since
`X\sim\mathcal N(0,1)`, you have


`F_X(x)=\mathbb P(X\le x)=\displaystyle\frac1{√(2\pi)}\int_(-\infty)^xe^(-t^2/2)\,\mathrm dt=\frac12+\frac12\mathrm{erf}\left(\frac x{\sqrt2}\right)`

(where
`\mathrm{erf}(x)` denotes the error function) and so


`F_Y(y)=\left(\frac12+\frac12\mathrm{erf}\left(√(\frac y2)\right)\right)-\left(\frac12+\frac12\mathrm{erf}\left(-√(\frac y2)\right)\right)`

`F_Y(y)=\mathrm{erf}\left(√(\frac y2)}\right)`

`\implies f_Y(y)=(\mathrm dF_Y(y))/(\mathrm dy)=\frac1{√(2\pi)}(e^(-y/2))/(\sqrt y)`

where
`y\ge0`.